derivative log_{4}(sec^2(5x^3-4x^2))

解答

导数

lo g_{4} (sec^{2} (5 x^{3} - 4 x^{2}))

\frac{tan ( 5 x ^{3} - 4 x ^{2} ) ( 15 x ^{2} - 8 x )}{ln ( 2 )}

求解步骤

\frac{d}{d x} (lo g_{4} (sec^{2} (5 x^{3} - 4 x^{2})))

使用对数计算法则:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( sec ^{2} ( 5 x ^{3} - 4 x ^{2} ) )}{ln ( 4 )})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (sec^{2} (5 x^{3} - 4 x^{2})))

使用连锁律:

\frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \frac{d}{d x} (sec^{2} (5 x^{3} - 4 x^{2}))

= \frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \frac{d}{d x} (sec^{2} (5 x^{3} - 4 x^{2}))

\frac{d}{d x} (sec^{2} (5 x^{3} - 4 x^{2})) = 2 sec^{2} (5 x^{3} - 4 x^{2}) tan (5 x^{3} - 4 x^{2}) (15 x^{2} - 8 x)

= \frac{1}{ln ( 4 )} \cdot \frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \cdot 2 sec^{2} (5 x^{3} - 4 x^{2}) tan (5 x^{3} - 4 x^{2}) (15 x^{2} - 8 x)

化简

\frac{1}{ln ( 4 )} \cdot \frac{1}{sec ^{2} ( 5 x ^{3} - 4 x ^{2} )} \cdot 2 sec^{2} (5 x^{3} - 4 x^{2}) tan (5 x^{3} - 4 x^{2}) (15 x^{2} - 8 x) : \frac{tan ( 5 x ^{3} - 4 x ^{2} ) ( 15 x ^{2} - 8 x )}{ln ( 2 )}

= \frac{tan ( 5 x ^{3} - 4 x ^{2} ) ( 15 x ^{2} - 8 x )}{ln ( 2 )}

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